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In the book of Cohn. Measure Theory 2nd edition there is the proposition 1.1.6:

Each closed subset of $\mathbb{R}^d$ is a $G_\delta$, and each open subset of $\mathbb{R}^d$ is an $F_\sigma$.

($G_\delta$ and $F_\sigma$ has the usual definitions, e.g. sets in $\mathcal{G}_\delta$ are called $G_\delta$'s, where $\mathcal{G}_\delta$ is the collection of all the intersections of sequences of sets in $\mathcal{G}$, with $\mathcal{G}$ the family of all open subsets of $\mathbb{R}^d$.)

Now it says: " It now follows (see proposition 1.1.6) that all the inclusions in Fig. 1.1 below are valid.

enter image description here

Well... I don't understand how the author uses the proposition in order to claim automatically that the figure follows from it. I can work out without the proposition proofs like $\mathcal{G}_\delta \subset\mathcal{F}_{\sigma\delta}$, but I don't see how to fit the prop. in e.g. showing $\mathcal{G}_{\delta\sigma}\subset\mathcal{F}_{\sigma\delta\sigma}$ because $\mathcal{G}_{\delta\sigma}$ contains open and closed subsets as well as $\mathcal{F}_{\sigma\delta}$, so why not $\mathcal{G}_{\delta\sigma}\subset\mathcal{F}_{\sigma\delta}$?

user2820579
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  • Hi there! I got the same question when I read the book. Thanks a lot for post it here. It really helps a lot! Just out of curiosity, how did you work it out without the proposition; for example, $\mathcal{G}{\delta} \subset \mathcal{F}{\sigma\delta}$ or $\mathcal{F}{\sigma} \subset \mathcal{G}{\delta\sigma}$? Do you mind adding it to your question? I really appreciate it! – Beerus Mar 11 '24 at 20:58

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Since $$\mathcal E_{\sigma}=\left\{\left.\cup_{n\in\mathbb N}A_n~\right|~(A_n)_{n\in\mathbb N} \subset\mathcal E\right\}$$ and $$\mathcal E_{\delta}=\left\{\left.\cap_{n\in\mathbb N} A_n~\right|~(A_n)_{n\in\mathbb N}\subset\mathcal E\right\},$$ whenever $\mathcal A\subset\mathcal B$, you have $\mathcal A_\sigma\subset\mathcal B_\sigma$ and $\mathcal A_\delta\subset\mathcal B_\delta.$

E.g. $\mathcal{G}_{\delta\sigma}\subset\mathcal{F}_{\sigma\delta\sigma},$ because $\mathcal{G}_\delta\subset\mathcal{F}_{\sigma\delta},$ because $\mathcal G\subset\mathcal{F}_\sigma$ by proposition 1.1.6.

But your conjecture $\mathcal{G}_{\delta\sigma}\subset\mathcal{F}_{\sigma\delta}$, equivalent to $\mathcal{G}_{\delta\sigma}=\mathcal{F}_{\sigma\delta}$, doesn't hold, since Borel hierarchy on $\mathbb R$ doesn't stabilize before $\omega_1$.

Anne Bauval
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